# Graph Potential Energy over Position

• merzperson
In summary, the problem involves finding the potential energy of a system using the equation F = -dU/dr, where F is the force and U is the potential energy. The change in potential energy is equal to the work done with a minus sign, and the change in kinetic energy is equal to the work done. This is only applicable for conservative forces, which can be derived from a potential.
merzperson

## Homework Statement

The figure shows the force Fx exerted on a particle that moves along the x-axis. Draw a graph of the particle's potential energy as a function of position x from 0m to 1.1m. Let U be zero at x=0m.

W = F*d
U = mgh

## The Attempt at a Solution

I'm pretty confused about how to start this problem. How do you know what equation to use in this situation?

My best idea is to use W = F*d because it includes both Force and Displacement, and relates them with Work, which is equal to the change in Energy of the system. However, I don't know how to use this to find the Potential Energy at a given Position. Please help!

You need the general relationship between potential energy and force.
U=mgh is just a special case and it has nothing to do with this problem.

Have you seen the relation $$F = - \frac{dU}{dr}$$ before?
If not, you probably have to review your textbook or lecture notes.
W = F.d only holds true for a constant force, which it clearly is not in this case.

Thanks guys!

Using the equation F = -dU/dr we can solve for dU:

dU = -Fdr

And then plug in the values for force (F) and displacement (dr). However, what do I plug in for the displacement? For example, at x=0.5

dU = -(2)(0.5) = -1

Is this correct? I'm not sure I'm using the displacement correctly. Also, once I find dU, how do I find U? Do I just take a small interval? Thank you for your help.

In order to find U you need to integrate the force over the position.
You can find F(x) from the graph (separately for the two sections) and then integrate, again for the two sections.

Hm...actually on second thought, I'm dubious as to whether the potential energy of the system actually changes. Assuming that we are not doing any work against forces like gravity, since the force acts along the direction of motion of the particle, the work done on the body goes solely to increasing its kinetic energy.

Of course there is a change in potential energy. Once again, gravitational potential energy is just one special case of potential energy.
The change in potential energy is equal to the the work done, with a minus sign.
The change in kinetic energy is equal to the work.
If the KE increases, the PE decreases.
If you need a special case to understand better, the force pictured here behaves like an elastic force (for each one of the two segments, not for the whole thing). Think about a ball attached to a spring and how the potential and kinetic energies vary during the motion.

nasu said:
Of course there is a change in potential energy. Once again, gravitational potential energy is just one special case of potential energy.
The change in potential energy is equal to the the work done, with a minus sign.
The change in kinetic energy is equal to the work.
If the KE increases, the PE decreases.
That I do not agree with. That the change in KE is equivalent to the change in PE magnitude-wise is only true for internal forces acting within the system of consideration. For a gravitational system of masses, or a spring-mass system, that clearly holds true. But what of situations such as a force pushing a block horizontally along the ground, or compression of a gas?

Thanks for all your help, I solved the problem. The reason for my confusion was because I never learned to integrate. I was doing the problem backwards.

Fightfish said:
That I do not agree with. That the change in KE is equivalent to the change in PE magnitude-wise is only true for internal forces acting within the system of consideration. For a gravitational system of masses, or a spring-mass system, that clearly holds true. But what of situations such as a force pushing a block horizontally along the ground, or compression of a gas?

If the force is conservative then it can be derived from a potential. The problem asks to calculate the potential energy so I assume this is a force that can be derived from a potential.
I mean for the problem at hand.

I did not mean and did not make a general statement. The kinetic/potential energy transformation is only valid for these cases when the potential energy can be defined and I think is reasonable to assume that this is the case in this problem.

It is interesting to think about the case of an object pulled along a horizontal surface, and accelerated.
Where does the kinetic energy comes from? It depends on where that force comes from but somewhere some other kind of energy must decrease, right?

## 1. What is graph potential energy over position?

Graph potential energy over position is a visual representation of how potential energy changes with the position of an object. It shows the relationship between potential energy and position and can be used to analyze the behavior of a system.

## 2. How is potential energy calculated in a graph of potential energy over position?

Potential energy is calculated by multiplying the force acting on an object by the distance the object moves. In a graph of potential energy over position, the height of the graph at a certain point represents the potential energy at that position.

## 3. What does a flat line on a graph of potential energy over position indicate?

A flat line on a graph of potential energy over position indicates that the potential energy does not change with position. This could mean that there is no force acting on the object, or that the force is perpendicular to the direction of motion.

## 4. How can a graph of potential energy over position be used to determine stable and unstable equilibrium?

In a graph of potential energy over position, stable equilibrium occurs when the graph has a minimum at a certain position, indicating that the object is in a stable state and will return to that position if displaced. Unstable equilibrium occurs when the graph has a maximum at a certain position, indicating that the object is in an unstable state and will move away from that position if displaced.

## 5. How does the shape of a graph of potential energy over position change with different types of forces?

The shape of a graph of potential energy over position can vary depending on the type of force acting on the object. For example, a linear graph with a positive slope indicates a constant force, while a curved graph indicates a non-constant force. The specific shape of the graph can also provide information about the strength and direction of the force.

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